Question

我有一个这样的数据框。

       Date       price     mid      std         top             btm
     ..............
    1999-07-21  8.6912  8.504580    0.084923    9.674425    8.334735
    1999-07-22  8.6978  8.508515    0.092034    8.692583    8.324447
    1999-07-23  8.8127  8.524605    0.118186    10.760976   8.288234
    1999-07-24  8.8779  8.688810    0.091124    8.871057    8.506563
     ..............

我想创建一个名为“ diff”的新列。如果连续输入'price'>'top'，那么我想用该行价格的几何平均收益率和前n-5行的价格的几何平均收益率来填充此行的'diff'（5天几何均值））。

例如，在1999-07-22行中，价格大于最高价，因此我想在此行中用07-22和07-17的几何平均值填充“ diff”（请注意，日期可能不是连续的因为假期不包括在内）。仅一小部分行满足需求。因此，“ diff”中的大多数值都将丢失值。

能否请您告诉我如何在python中做到这一点？

Answer 1

将Series.diff与Series.where一起用于集合NaN：

df['diff'] = df['price'].diff().where(df['price'] > df['top'])
print (df)
             price       mid       std        top       btm    diff
Date                                                               
1999-07-21  8.6912  8.504580  0.084923   9.674425  8.334735     NaN
1999-07-22  8.6978  8.508515  0.092034   8.692583  8.324447  0.0066
1999-07-23  8.8127  8.524605  0.118186  10.760976  8.288234     NaN
1999-07-24  8.8779  8.688810  0.091124   8.871057  8.506563  0.0652

编辑：

我相信您需要：

df['Date'] = pd.to_datetime(df['Date'])
df = df.set_index('Date')

from scipy.stats.mstats import gmean

df['gmean'] = (df['price'].rolling('5d')
                          .apply(gmean, raw=True)
                          .where(df['price'] > df['top']))
print (df)
             price       mid       std        top       btm     gmean
Date                                                                 
1999-07-21  8.6912  8.504580  0.084923   9.674425  8.334735       NaN
1999-07-22  8.6978  8.508515  0.092034   8.692583  8.324447  8.694499
1999-07-23  8.8127  8.524605  0.118186  10.760976  8.288234       NaN
1999-07-24  8.8779  8.688810  0.091124   8.871057  8.506563  8.769546

Answer 2

您可以通过获取price和top列的差值，然后将<= 0的那些值分配为NaN或零值来实现：

import pandas as pd
import numpy as np

df = pd.DataFrame(...)

df['diff'] = df['price'] - df['top']

df.loc[df['diff'] <= 0, 'diff'] = np.NaN # or 0

Answer 3

这是另一种解决方案：

import pandas as pd
from functools import reduce

__name__ = 'RunScript'

ddict = {
    'Date':['1999-07-21','1999-07-22','1999-07-23','1999-07-24',],
    'price':[8.6912,8.6978,8.8127,8.8779],
    'mid':[8.504580,8.508515,8.524605,8.688810],
    'std':[0.084923,0.092034,0.118186,0.091124],
    'top':[9.674425,8.692583,10.760976,8.871057],
    'btm':[8.334735,8.324447,8.288234,8.506563],
    }


data = pd.DataFrame(ddict)


def geo_mean(iter):
    """
        Geometric mean function. Pass iterable
    """
    return reduce(lambda a, b: a * b, iter) ** (1.0 / len(iter))


def set_geo_mean(df):
    # Shift the price row down one period
    data['shifted price'] = data['price'].shift(periods=1)

    # Create a masked expression that evaluates price vs top
    masked_expression = df['price'] > df['top']

    # Return rows from dataframe where masked expression is true
    masked_data = df[masked_expression]

    # Apply our function to the relevant rows
    df.loc[masked_expression, 'geo_mean'] = geo_mean([masked_data['price'], masked_data['shifted price']])

    # Drop the shifted price data column once complete
    df.drop('shifted price', axis=1, inplace=True)


if __name__ == 'RunScript':
    # Call function and pass dataframe argument.
    set_geo_mean(data)

计算特定行的几何平均收益

3 个答案: